Problem: Simplify and expand the following expression: $ \dfrac{1}{4y - 24}- \dfrac{5}{4y - 8}+ \dfrac{4y}{y^2 - 8y + 12} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{1}{4y - 24} = \dfrac{1}{4(y - 6)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4y - 8} = \dfrac{5}{4(y - 2)}$ We can factor the quadratic in the third term: $ \dfrac{4y}{y^2 - 8y + 12} = \dfrac{4y}{(y - 6)(y - 2)}$ Now we have: $ \dfrac{1}{4(y - 6)}- \dfrac{5}{4(y - 2)}+ \dfrac{4y}{(y - 6)(y - 2)} $ The least common multiple of the denominators is: $ 16(y - 6)(y - 2)$ In order to get the first term over $16(y - 6)(y - 2)$ , multiply by $\dfrac{4(y - 2)}{4(y - 2)}$ $ \dfrac{1}{4(y - 6)} \times \dfrac{4(y - 2)}{4(y - 2)} = \dfrac{4(y - 2)}{16(y - 6)(y - 2)} $ In order to get the second term over $16(y - 6)(y - 2)$ , multiply by $\dfrac{4(y - 6)}{4(y - 6)}$ $ \dfrac{5}{4(y - 2)} \times \dfrac{4(y - 6)}{4(y - 6)} = \dfrac{20(y - 6)}{16(y - 6)(y - 2)} $ In order to get the third term over $16(y - 6)(y - 2)$ , multiply by $\dfrac{16}{16}$ $ \dfrac{4y}{(y - 6)(y - 2)} \times \dfrac{16}{16} = \dfrac{64y}{16(y - 6)(y - 2)} $ Now we have: $ \dfrac{4(y - 2)}{16(y - 6)(y - 2)} - \dfrac{20(y - 6)}{16(y - 6)(y - 2)} + \dfrac{64y}{16(y - 6)(y - 2)} $ $ = \dfrac{ 4(y - 2) - 20(y - 6) + 64y} {16(y - 6)(y - 2)} $ Expand: $ = \dfrac{4y - 8 - 20y + 120 + 64y}{16y^2 - 128y + 192} $ $ = \dfrac{48y + 112}{16y^2 - 128y + 192}$ Simplify: $ = \dfrac{3y + 7}{y^2 - 8y + 12}$